Trigonometry Examples

$2 i^{2} + 3 i^{3}$

Step 1

Simplify each term.

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Step 1.1

Rewrite $i^{2}$ as $- 1$ .

$2 \cdot - 1 + 3 i^{3}$

Step 1.2

Multiply $2$ by $- 1$ .

$- 2 + 3 i^{3}$

Step 1.3

Factor out $i^{2}$ .

$- 2 + 3 (i^{2} \cdot i)$

Step 1.4

Rewrite $i^{2}$ as $- 1$ .

$- 2 + 3 (- 1 \cdot i)$

Step 1.5

Rewrite $- 1 i$ as $- i$ .

$- 2 + 3 (- i)$

Step 1.6

Multiply $- 1$ by $3$ .

$- 2 - 3 i$

Step 2

This is the trigonometric form of a complex number where $| z |$ is the modulus and $θ$ is the angle created on the complex plane.

$z = a + b i = | z | (\cos (θ) + i \sin (θ))$

Step 3

The modulus of a complex number is the distance from the origin on the complex plane.

$| z | = \sqrt{a^{2} + b^{2}}$ where $z = a + b i$

Step 4

Substitute the actual values of $a = - 2$ and $b = - 3$ .

$| z | = \sqrt{{(- 3)}^{2} + {(- 2)}^{2}}$

Step 5

Find $| z |$ .

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Step 5.1

Raise $- 3$ to the power of $2$ .

$| z | = \sqrt{9 + {(- 2)}^{2}}$

Step 5.2

Raise $- 2$ to the power of $2$ .

$| z | = \sqrt{9 + 4}$

Step 5.3

Add $9$ and $4$ .

$| z | = \sqrt{13}$

Step 6

The angle of the point on the complex plane is the inverse tangent of the complex portion over the real portion.

$θ = \arctan (\frac{- 3}{- 2})$

Step 7

Since inverse tangent of $\frac{- 3}{- 2}$ produces an angle in the third quadrant, the value of the angle is $4.12438637$ .

$θ = 4.12438637$

Step 8

Substitute the values of $θ = 4.12438637$ and $| z | = \sqrt{13}$ .

$\sqrt{13} (\cos (4.12438637) + i \sin (4.12438637))$